betancurj wrote:
Hi Bunuel,
This is my first time participating in gmatclub. Firs of all thank you for always posting stuff that really helps and for always giving a good solution to math problems.
I've been struggling trying to find out an easy procedure for this question. Can you help?
Thanks again
Juan
Let me try to answer.
You are given that the standard equation of a circle ---> (x−h)^2+(y−k)^2=r^2 where, r=radius of the circle.
Now, you are given another circle with the equation ---> x^2 + y^2 - 8x + 2y = -1 , when you compare this to the standard equation of the circle, you should realize that it would be straightforward to calculate the radius of the circle if you can express x^2 + y^2 - 8x + 2y = -1 in the form of (x−h)^2+(y−k)^2=r^2
In order to make complete squares similar to (x-h)^2 or (y-k)^2, you need to remember that (x+a)^2=x^2+a^2+2ax and (x-a)^2=x^2+a^2-2ax
With these relations in mind, lets go back to x^2 + y^2 - 8x + 2y = -1 with a focus on converting x^2 + y^2 - 8x + 2y to 2 complete squares (1 with x and 1 with y).
x^2 + y^2 - 8x + 2y ---> (x^2-8x+16-16)+(y^2+2y+1-1) = (x-4)^2 -16 + (y+1)^2-1 ---> x^2 + y^2 - 8x + 2y = -1 can thus be written as :
(x-4)^2 -16 + (y+1)^2-1 = -1 --->(x-4)^2 + (y+1)^2 = 16 = 4^2.
NOw you can compare this form to (x−h)^2+(y−k)^2=r^2 to get h = 4, k=-1 and r=radius = 4.
Thus, D is the correct answer.
For questions such as these you need to move backwards from what has been given to you and to modify the given values into a more usable form.
Hope this helps.